7. A one-tailed hypothesis test with the t statistic Antisocial personality diso
ID: 3062501 • Letter: 7
Question
7. A one-tailed hypothesis test with the t statistic Antisocial personality disorder (ASPD) is characterized by deceitfulness, reckless disregard for the well-being of others, a diminished capacity for remorse, superficial charm, thrill seeking, and poor behavioral control. ASPD is not normally diagnosed in children or adolescents, but antisocial tendencies can sometimes be recognized in childhood or early adolescence. James Blair and his colleagues have studied the ability of children with antisocial tendencies to recognize facial expressions that depict sadness, happiness, anger, disgust, fear, and surprise. They have found that children with antisocial tendencies have selective impairments, with significantly more difficulty recognizing fearful and sad expressions. Suppose you have a sample of 30 children with antisocial tendencies and you are particularly interested in the emotion of disgust. You find that while the average 14-year-old has a score of 8.95 on the emotion recognition scale, your sample of 14-year-old children with antisocial tendencies has an average score of 9.65 with a standard deviation of 3.47. A higher score represents that it took a higher degree of emotional expression for the child to correctly identify the emotion of disgust, indicating greater difficulty recognizing the emotion. Assume that scores on the emotion recognition scale are normally distributed. The null hypothesis is that your sample of children with antisocial tendencies would have no more difficulty recognizing emotion than the general population of 14-year-olds. Stated using symbols This is a tailed test. Given what you know, you will evaluate this hypothesis using a statistic.Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: < 8.95
Alternative hypothesis: > 8.95
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = s / sqrt(n)
S.E = 0.6335
DF = n - 1
D.F = 29
t = (x - ) / SE
t = 1.10
tcritical = 1.699
Critical region is t > 1.699
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
The observed sample mean produced a t statistic test statistic of 1.10
Interpret results. Since the t-value (1.10) does not lie in the critical region therefore, we cannot reject the null hypothesis.
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